AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.1 Question 1.
What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:

 Number Square of the units digIt Units digit of a squared number i) 39 92 = 9 x 9 = 81 1 ii) 297 72 = 7 x 7 = 49 9 Iii) 5125 52 = 5 x 5 = 25 5 iv) 7286 62 = 6 x 6 = 36 6 v) 8742 22 = 2 x 2 = 4 4 Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256
(iv) 321
(v) 600
Solution:

 Number Prime factorizatlon Perfect square numbers Yes/No i) 121 121 = 11 x 11 = 112 yes ii) 136 136 = 8 x 17 = 2 x 2 x 2 x 17 No iii) 256 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 28 = (24)2 yes iv) 321 321 = 3 x 107 No v) 600 600 = 120 x 5 = 12 x 10 x 5 = 2 x 2 x 2 x 3 x 5 x 5 No

Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:

 Number Units digit of square of a number Even / Odd (i) 431 12 = 1 1 , odd (ii) 2826 62 = 36 6, odd (iii) 8204 42 = 16 6 , even (iv)17779 92 = 81 1 , odd (v) 99998 82 = 64 4 , even

Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214 Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n2]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 92 = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 132 = 13 x 13 = 169